MCQ
Given

  Reaction    Energy Change (in $kJ$ )
  $Li(s) \to  Li(g)$    $161$
  $Li(g) \to  Li^+(g)$    $520$
  $\frac {1}{2}F_2(g)\,\to F(g)$    $77$
  $F(g) + e^- \to F^-(g)$   (Electron gain enthalpy)
  $Li^+ (g) + F^-(g) \to  LiF(s)$    $-1047$
  $Li (s) + \frac {1}{2}F_2(g)\to LiF(s)$    $-617$

Based on data provided, the value of electron gain enthalpy of fluorine would be.....$kJ\,mol^{-1}$

  • A
    $-300$
  • B
    $-350$
  • $-328$
  • D
    $-228$

Answer

Correct option: C.
$-328$
c
Applying Hess's Law 

${\Delta _f}{H^o} = {\Delta _{sub}}H + \frac{1}{2}{\Delta _{diss}}H + I.E. + E.A + {\Delta _{lattice}}H$

${\Delta _f}{H^o} = {\Delta _{sub}}H + \frac{1}{2}{\Delta _{diss}}H + I.E. + E.A + {\Delta _{lattice}}H$

$E.A. =  - 617 + 286 =  - 328\,kJ\,mo{l^{ - 1}}$

$\therefore $ electron affinity of fluorine

$ =  - 328\,kJ\,mo{l^{ - 1}}$

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